Question

Find the zeros of the polynomial $f\left(x\right)=x^2-2$ and verify the relationship between its zeros and coefficients.

Answer 1

We have,

$f\left(x\right)=x^2-2$

$=x^2-(\sqrt{2}{)}^2$

$=(x-\sqrt{2})(x+\sqrt{2})$

The zeroes of$f\left(x\right)$ are given by $f\left(x\right)=0$

$(x-\sqrt{2})(x+\sqrt{2})=0$

$(x-\sqrt{2})=0$ or$,(x+\sqrt{2})=0$

$x=\sqrt{2}$ or $x=-\sqrt{2}$

Thus ,the zeroes of $f\left(x\right)$ are $\alpha =\sqrt{2}$ and $\beta =-\sqrt{2}$

Now,

Sum of the zeroes$=\alpha +\beta =\sqrt{2}+(-\sqrt{2})$

$=0$

and, $-\left(\frac{coefficientofx}{coefficientof{x}^2}\right)=-\left(\frac{0}{1}\right)=0$

Therefore sum of the zeroes$=-\left(\frac{coefficientofx}{coefficientof{x}^2}\right)$

Product of the zeroes$=\alpha \times \beta =\sqrt{2}\times -\sqrt{2}=-2$

and,$\frac{constantterm}{coefficientof{x}^2}=\frac{-2}{1}=-2$

Therefore, product of zeros $=\frac{constantterm}{coefficientof{x}^2}$